\(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx\) [415]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=-\frac {2^{\frac {1}{2}+m} a c \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m}}{3 f} \] Output:

-1/3*2^(1/2+m)*a*c*cos(f*x+e)^3*hypergeom([3/2, 1/2-m],[5/2],1/2-1/2*sin(f 
*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^(-1+m)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.16 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.60 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=-\frac {2^{-m} c \left (i+e^{i (e+f x)}\right ) \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^m \left (-\left ((-1+m) m \operatorname {Hypergeometric2F1}\left (1,m,-m,i e^{i (e+f x)}\right )\right )+e^{i (e+f x)} (1+m) \left (-2 i (-1+m) \operatorname {Hypergeometric2F1}\left (1,1+m,1-m,i e^{i (e+f x)}\right )+e^{i (e+f x)} m \operatorname {Hypergeometric2F1}\left (1,2+m,2-m,i e^{i (e+f x)}\right )\right )\right ) (-1+\sin (e+f x))}{\left (-i+e^{i (e+f x)}\right )^2 f (-1+m) m (1+m)} \] Input:

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]),x]
 

Output:

-((c*(I + E^(I*(e + f*x)))*(((-I)*a*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x 
)))^m*(-((-1 + m)*m*Hypergeometric2F1[1, m, -m, I*E^(I*(e + f*x))]) + E^(I 
*(e + f*x))*(1 + m)*((-2*I)*(-1 + m)*Hypergeometric2F1[1, 1 + m, 1 - m, I* 
E^(I*(e + f*x))] + E^(I*(e + f*x))*m*Hypergeometric2F1[1, 2 + m, 2 - m, I* 
E^(I*(e + f*x))]))*(-1 + Sin[e + f*x]))/(2^m*(-I + E^(I*(e + f*x)))^2*f*(- 
1 + m)*m*(1 + m)))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3215, 3042, 3168, 80, 79}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (c-c \sin (e+f x)) (a \sin (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c-c \sin (e+f x)) (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3215

\(\displaystyle a c \int \cos ^2(e+f x) (\sin (e+f x) a+a)^{m-1}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \cos (e+f x)^2 (\sin (e+f x) a+a)^{m-1}dx\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^3 c \cos ^3(e+f x) \int \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^3 c 2^{m-\frac {1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-2} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}} \sqrt {a-a \sin (e+f x)}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {a^2 c 2^{m+\frac {1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 f}\)

Input:

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]),x]
 

Output:

-1/3*(2^(1/2 + m)*a^2*c*Cos[e + f*x]^3*Hypergeometric2F1[3/2, 1/2 - m, 5/2 
, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^ 
(-2 + m))/f
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [F]

\[\int \left (a +\sin \left (f x +e \right ) a \right )^{m} \left (c -c \sin \left (f x +e \right )\right )d x\]

Input:

int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e)),x)
 

Output:

int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e)),x)
 

Fricas [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="fricas")
 

Output:

integral(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
 

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=- c \left (\int \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- \left (a \sin {\left (e + f x \right )} + a\right )^{m}\right )\, dx\right ) \] Input:

integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e)),x)
 

Output:

-c*(Integral((a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(-(a*sin(e 
 + f*x) + a)**m, x))
 

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="maxima")
 

Output:

-integrate((c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
 

Giac [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="giac")
 

Output:

integrate(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c-c\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:

int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)),x)
 

Output:

int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)), x)
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=c \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m}d x -\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right )\right ) \] Input:

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x)
 

Output:

c*(int((sin(e + f*x)*a + a)**m,x) - int((sin(e + f*x)*a + a)**m*sin(e + f* 
x),x))